Question

The number of integral terms in the expansion of ${\left(5^{\frac{1}{6}}+7^{\frac{1}{9}}\right)}^{1824}$ is

• A. $84$
• B. $96$
• C. $91$
• D. $102$

Answer 1

Answer: (D)

$T_{r+1}{=}^{1824}{C}_r{\left(5^{\frac{1}{6}}\right)}^{1824-r}{\left(7^{\frac{1}{9}}\right)}^r$
${=}^{1824}{C}_r\left(5^{\frac{1824-r}{6}}\right)\left(7^{\frac{r}{9}}\right)$
${=}^{1824}{C}_r\left(5^{304-\frac{r}{6}}\right)\left(7^{\frac{r}{9}}\right)$
Both $\frac{r}{6}$ and $\frac{r}{9}$ are integers, if $r$ is a multiple of $18$
$\Rightarrow r=0,18,36,......,1818$
Let, the number of terms be $n$
$\Rightarrow 1818=0+\left(n-1\right)18\Rightarrow n=102.$